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2 years ago

The product of the roots of 8x2 - 2x = 1 is: -1/8 1/8

Mathematics

2 answers:

Ivahew [28]2 years ago

4 0

Answer:

1/8

Step-by-step explanation:

ioda2 years ago

4 0

Answer:

-1/8

Step-by-step explanation:

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Heeeeeeeeeeeeeeeeelp

lys-0071 [83]

Answer:

a b

Step-by-step explanation:

7 0

3 years ago

A dog walker charges a flat rate of $6 per walk plus an hourly rate of $30. How much does the dog walker charge for a 45 minute

zvonat [6]

Step 1

Find the equation in function notation

Let

h-------> the number of hours

y-------> the function for the walker fee in dollars

we know that

the hourly rate is $30\frac{\$}{hour}$

$y=6+30h$

in this linear equation

the independent variable is the variable h

the dependent variable is the variable y

Convert to function notation

Let

$f(h)=y$

$f(h)=6+30h$

Step 2

Find how much does the dog walker charge for a 45 minute walk

Convert the time in hours

$1\ hour=60\ minutes$

$45\ minutes=45/60=0.75\ hours$

substitute in the equation

For $h=0.75\ hours$

$f(0.75)=6+30*(0.75)=\$28.5$

therefore

the answer is

the dog walker charge for a 45 minute walk $\$28.5$

Statements

Step 3

The dependent variable is the number of hours true or false

The statement is false,

because the independent variable is the number of hours and the dependent variable is the walker fee in dollars

Step 4

The function for the walker fee is f(h)= 30h+6 true or false

The statement is True --------> see the Step 1

Step 5

The dog walker charges $22.5 for a 45 min walk true or false

The statement is False

Because the dog walker charges for a 45 minute walk $\$28.5$

5 0

3 years ago

HELPPP PLEASE WILL MARK BRAINLEST IF RIGHT!!!

Sholpan [36]

Answer:

C i think

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Given the function f(x) = x^4 + 3x^3 - 2x^2 - 6x - 1, use intermediate theorem to decide which of the following intervals contai

marta [7]

$f(x) = x^4 + 3x^3 - 2x^2 - 6x - 1$

Lets check with every option

(a) [-4,-3]

We plug in -4 for x and -3 for x

$f(-4) = (-4)^4 + 3(-4)^3 - 2(-4)^2 - 6(-4) - 1= 55$

$f(-3) = (-3)^4 + 3(-3)^3 - 2(-3)^2 - 6(-3) - 1= -1$

f(-4) is positive and f(-3) is negative. there is some value at x=c on the interval [-4,-3] where f(c)=0. so there exists atleast one zero on this interval.

(b) [-3,-2]

We plug in -3 for x and -2 for x

$f(-3) = (-3)^4 + 3(-3)^3 - 2(-3)^2 - 6(-3) - 1= -1$

$f(-2) = (-2)^4 + 3(-2)^3 - 2(-2)^2 - 6(-2) - 1= -5$

f(-2) is negative and f(-3) is negative. there is no value at x=c on the interval [-3,-2] where f(c)=0.

(c) [-2,-1]

We plug in -2 for x and -1 for x

$f(-2) = (-2)^4 + 3(-2)^3 - 2(-2)^2 - 6(-2) - 1= -5$

$f(-1) = (-1)^4 + 3(-1)^3 - 2(-1)^2 - 6(-1) - 1= 1$

f(-2) is negative and f(-1) is positive. there is some value at x=c on the interval [-2,-1] where f(c)=0. so there exists atleast one zero on this interval.

(d) [-1,0]

We plug in -1 for x and 0 for x

$f(-1) = (-1)^4 + 3(-1)^3 - 2(-1)^2 - 6(-1) - 1= 1$

$f(0) = (0)^4 + 3(0)^3 - 2(0)^2 - 6(0) - 1= -1$

f(-1) is positive and f(0) is negative. there is some value at x=c on the interval [-1,0] where f(c)=0. so there exists atleast one zero on this interval.

(e) [0,1]

We plug in 0 for x and 1 for x

$f(0) = (0)^4 + 3(0)^3 - 2(0)^2 - 6(0) - 1= -1$

$f(1) = (1)^4 + 3(1)^3 - 2(1)^2 - 6(1) - 1= -5$

f(0) is negative and f(1) is negative. there is no value at x=c on the interval [0,1] where f(c)=0.

(f) [1,2]

We plug in 1 for x and 2 for x

$f(1) = (1)^4 + 3(1)^3 - 2(1)^2 - 6(1) - 1= -5$

$f(2) = (2)^4 + 3(2)^3 - 2(2)^2 - 6(2) - 1= 19$

f(-4) is positive and f(-3) is negative. there is some value at x=c on the interval [-4,-3] where f(c)=0. so there exists atleast one zero on this interval.

so answers are (a) [-4,-3], (c) [-2,-1], (d) [-1,0], (f) [1,2]

8 0

3 years ago

Read 2 more answers

Common factor for 12 and 24

Afina-wow [57]

2 is one, i really don’t remember this. I think 6 and 4 3 i think.

8 0

3 years ago